
\prob{003F}{矩形翻折}

\begin{figure}[htbp]
  \centering
  \image{003F}
  \caption{003F：矩形翻折} \label{fig:003F}
\end{figure}

如图~\ref{fig:003F}，在矩形$ABCD$中，$AB = 4, BC = 3$，$P$为边$BC$上一点，将$\triangle CDP$沿$DP$折叠，点$C$落在点$E$处，$DE, PE$分别交$AB$于点$F, G$。若$EG = BG$，求$BF$的长。
\problabels{yellow/平面几何, green/长度问题}

\ans{$BF = 12/5$}

\subsection{勾股定理} \label{subsec:003F-pyth}

设$BF = x$。

\begin{align*}
  \because  {}& \text{$\triangle EDP$ 由 $\triangle CDP$ 折叠而来} \\
  \therefore{}& \angle C = \angle E, PC = PE, CD = DE \\
  \because  {}& \text{四边形 $ABCD$ 为矩形} \\
  \therefore{}& \angle A = \angle B = \angle C = 90^\circ, \\
  & AB = CD, BC = AD \\
  \therefore{}& \angle B = \angle E \\
  \because  {}& \text{在 $\triangle EFG$ 与 $\triangle BPG$ 中} \\
  & \begin{cases}
    \angle E = \angle B \\
    EG = BG \\
    \angle EGF = \angle BGP
  \end{cases} \\
  \therefore{}& \triangle EFG \cong \triangle BPG \\
  \therefore{}& FG = PG, EF = BP \\
  \because  {}& BF = BG + FG \\
  \therefore{}& BF = EG + PG = PE \\
  \because  {}& BF = x \\
  \therefore{}& BF = PE = PC = x \\
  \because  {}& BC = 3 \\
  \therefore{}& AD = 3, BP = 3 - x \\
  \therefore{}& EF = 3 - x \\
  \because  {}& AB = 4 \\
  \therefore{}& CD = 4, AF = AB - BF = 4 - x \\
  \therefore{}& DE = 4 \\
  \therefore{}& DF = DE - EF = x + 1 \\
  \text{又}\because{}& \angle A = 90^\circ \\
  \therefore{}& AD^2 + AF^2 = DF^2 \\
  \therefore{}& 3^2 + (4 - x)^2 = (x + 1)^2 \\
  \therefore{}& x = \frac{12}5 \\
  \therefore{}& BF = \frac{12}5
\end{align*}

综上，$BF$的长为$12/5$。

\subsection{面积法}

\emph{YCY提供的方法。}

根据解法\nameref{subsec:003F-pyth}知，$BF = PC$，设$BF = PC = x$，则易知$S_{\triangle CDP} = 2x$，$S_{BCDF} = 3(x + 4)/2 = 3x/2 + 6$。

又根据解法\nameref{subsec:003F-pyth}知

\[ \triangle EFG \cong \triangle BPG \]

因此

\begin{align*}
  S_{\triangle EFG} &= S_{\triangle BPG} \\
  S_{\triangle EFG} + S_{CDFP} &= S_{\triangle BPG} + S_{CDFP} \\
  S_{CDEP} &= S_{BCDF} \\
  2S_{\triangle CDP} &= S_{BCDF} \\
  4x &= \frac32x + 6 \\
  x &= \frac{12}5 \\
  BF &= \frac{12}5
\end{align*}

综上，$BF$的长为$12/5$。
